Re: [PHP-DB] Newbie Question $2

[email protected] ("Ethan Rosenberg, PhD") Wed, 18 Jun 2014 00:31:43 -0400
Newsgroups	php.db
Organization	Hygeia Biomedical Research, Inc.
Message-ID	<[email protected]>
On 06/17/2014 12:02 PM, [email protected] wrote:
> Hi Ethan,
>
> Here are some things to clean up your code:
>
> Your line:
>
> $phn = $_POST[phone];
>
> should use quotations as follows:
>
> $phn = $_POST['phone'];
>
> Your line:
>
> $sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
>
> Should use double quotes if you need the variable to be interpreted:
>
> $sql1 ="select Lname, Fname from Customers where Phone = $Phn ";
>
> Lastly, as people have mentioned PDO is probably the best way to go. Try connecting to your database with PDO. Look on Google for "PDO prepared statements" and use those instead of the mysql escape string method.
>
> Hope this helps,
>
> -Kevin
>
> Sent from Yahoo Mail on Android
>
>
IT WORKS!!!

Here is the code -

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<html>
<?php
   $bla = 1;
?>
     <head>
     </head>
         <body>
			<div align="center">
             <form method="post">
             <input type='text' name=phone></input>
             <input type='submit'>
			<br /><br /><br />
             </form>
			</div>
<?php
			error_reporting(-1);
			require '/home/ethan/PHP/ethan.inc';
			$db = "Store";
			$cxn = mysqli_connect($host,$user,$password,$db);

			$phn = $_POST[phone];
			$phn = (string)$phn;
			$dsh = '-';
			$Phn = 
$phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
			$sql1 ="select Lname, Fname from Customers where Phone = '$Phn' ";
			$result1 = mysqli_query($cxn, $sql1);
			if(!$result)
			{
?>				
				<div align="center">
				
				<strong>No Match Found</strong>
				<br /><br />
				</div>
<?php
			}
		
?>
			<div align="center">
			<table border="4" cellpadding="5" cellspacing="55" rules="all" 
frame="box">
			<tr class='heading'>
			<th>Last Name</th>
			<th>First Name</th>
<?php

             while($row1 = mysqli_fetch_row($result1))
             {

				$Lname     = $row1[0];
				$Fname     = $row1[1];



?>              <tr>
				<td> <?php echo $Lname; ?> </td>
				<td> <?php echo $Fname; ?> </td>
                 </tr>
<?php
               }
?>
               </table>
			</div>>
		</body>
</html>
	
As you [those that replied] accurately noted, the problem was with the 
quoting.

I appreciate all your comments, take them seriously and will use the 
information contained in them for future programming.

No matter how much skill in programming I have, I will remain a NEWBIE; 
ie, someone who wishes to grrow in knowledge and acknowledges that there 
are many programmers much more skilled than I.

Thanks again.

Ethan